Problem: Find $\lim_{h\to 0}\dfrac{5\arcsin\left(\dfrac{2}{3}+h\right)-5\arcsin\left(\dfrac{2}{3}\right)}{h}$. Choose 1 answer: Choose 1 answer: (Choice A) A $3\sqrt{5}$ (Choice B) B $\dfrac{\sqrt{5}}{3}$ (Choice C) C $\dfrac{3\sqrt{5}}{5}$ (Choice D) D The limit doesn't exist
Solution: The limit expression has the following form: $\lim_{h\to 0}\dfrac{f(t+h)-f(t)}{h}$ Therefore, it describes a derivative of a certain function $f$ at a certain $x$ -value $t$. Can you recognize what $f$ and $t$ are? Since the numerator is $5\arcsin\left(\dfrac{2}{3}+h\right)-5\arcsin\left(\dfrac{2}{3}\right)$, we can tell that the function is $f(x)=5\arcsin(x)$ and the $x$ -value is $\dfrac{2}{3}$. In other words, the limit expression is equal to $f'\left(\dfrac{2}{3}\right)$ for $f(x)=5\arcsin(x)$. Let's find $f'(x)$ : $f'(x)=5\cdot \dfrac{1}{\sqrt{1-x^2}}$ Now let's evaluate $f'\left(\dfrac{2}{3}\right)$ : $\begin{aligned}f'\left(\dfrac{2}{3}\right)&=5\cdot \dfrac{1}{\sqrt{1-\left(\dfrac23\right)^2}} \\\\ &=\dfrac{5}{\sqrt{\dfrac{5}{9}}} \\\\ &=\dfrac{5}{\left(\dfrac{\sqrt{5}}{3}\right)} \\\\ &=\dfrac{15}{\sqrt{5}} \\\\ &=3\sqrt{5}\end{aligned}$ In conclusion, $\lim_{h\to 0}\dfrac{5\arcsin\left(\dfrac{2}{3}+h\right)-5\arcsin\left(\dfrac{2}{3}\right)}{h}=3\sqrt{5}$.